## anonymous 3 years ago Trig identity.. I don't understand this one at all

1. anonymous

how do i make the left side into one big fraction

2. anonymous

$\huge \frac{ 1 }{ sinx+1 } - \frac{ 1 }{ sinx-1 } = \frac{ 2 }{ 1-\sin^2x}$

3. anonymous

typed it wrong wooops

4. anonymous

get common denominators on the left side: $$\huge \frac{ 1 }{ sinx+1 }\cdot (\color {red}{\frac{ sinx-1 }{ sinx-1 }}) - \frac{ 1 }{ sinx-1 }\cdot( \color {red}{\frac{ sinx+1 }{ sinx+1 }}) = \frac{ 2 }{ 1-\sin^2x}$$

5. anonymous

so the left side in simplified form will be: $$\huge \frac{(sinx-1)-(sinx+1)}{sin^2x-1}$$ can you simplify this?

6. anonymous

no that's where i get stuck how would i subtract the numerator

7. anonymous

just use distributive property to get rid of the parenthesis...

8. anonymous

@dpaInc i get this $\huge \frac{ \sin^2x+sinx-sinx-1 }{ sinx^2-1 }$

9. anonymous

how did you get that? all you're doing is subtracting....: $$\huge \frac{(sinx-1)-(sinx+1)}{sin^2x-1}$$ =$$\huge \frac{1\cdot (sinx-1)-1\cdot (sinx+1)}{sin^2x-1}$$ =$$\huge \frac{sinx-1-sinx-1}{sin^2x-1}$$ =$$\huge \frac{-2}{sin^2x-1}$$ =$$\huge \frac{2}{1-sin^2x}$$

10. anonymous

sinx-1-sinx-1 = -2 is that a trig rule or what ? :S